At a glance
足るを知りたい 配当120万円への道 ranks #4,677,833 globally for subscribers (180), #4,221,515 globally for views (42.5K), 58 videos published from JP.
Subscribers
180
Views
42,533
Videos
58
Charts
Subscribers, views & videos over time
Subscribers gained (weekly) · 6 mo
Views gained (weekly) · 6 mo
Engagement
Avg. views per video
733
Total views ÷ total videos
Views per subscriber
236×
Total views ÷ subscribers
Subs per video
3
Subscribers ÷ total videos
Analytics tables
About
30代共働き・子ども2人(3歳、0歳) 子どもの教育費用を「配当金」で準備することを目標に投資しています。 目標:年間配当120万円 現在:207,023円(2025年12月末時点) 毎月 ・家計簿公開 ・資産額公開 ・NISA運用 ・高配当投資の進捗 を記録しています。 教育費と老後資金をコツコツ準備するリアルを発信。 チャンネル登録して、一緒に見守っていただけたら嬉しいです。
- Handle
- @足るを知りたい
- Joined
- January 2, 2024
More features coming
We're working on new analytics and tools. Stay tuned.
More features coming
We're working on new analytics and tools. Stay tuned.
Frequently asked questions
How accurate are these YouTube channel analytics?
Subscriber, view, and video counts come from YouTube's public API and refresh regularly. Growth and revenue figures are estimates built from tracked historical data, so they can differ slightly from what a channel owner sees privately in YouTube Studio.
How is a channel's global and country rank calculated?
Rank compares this channel's subscriber count and view count against every other channel Socialcounts.org tracks, both worldwide and within its home country.
How is the revenue estimate calculated?
It's a rough public estimate based on a $0.7–$2.0 CPM (revenue per 1,000 views) applied to view counts — not actual creator earnings, which depend on ad rates, audience geography, and monetization settings YouTube doesn't expose publicly.
Can I check analytics for any YouTube channel, not just my own?
Yes — search any public YouTube channel by name, handle, or channel URL. No login is required to see the basics.
Is this YouTube channel analytics checker free?
Yes. You get 10 free channel and video analytics lookups per rolling hour without an account — logging in, which is also free, removes that limit entirely.